Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.4 Exercises - Page 288: 55

Answer

Average value $=\displaystyle \frac{2}{\pi}$ x$\approx$0,69, x$\approx$2.451

Work Step by Step

If $f$ is integrable on the closed interval $[a, b]$, then the average value of $f$ on the interval is $\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx$. ----------------- Average value = $\displaystyle \frac{1}{\pi-0}\int_{0}^{\pi}\sin xdx$ $=[-\displaystyle \frac{1}{\pi}\cos x]_{0}^{\pi}$ $=-\displaystyle \frac{1}{\pi}(\cos\pi-\cos 0)$ $=-\displaystyle \frac{1}{\pi}(-1-1)$ $=\displaystyle \frac{2}{\pi}$ To find x where f(x)=average value, solve (on [0,$\pi$] ) $\displaystyle \sin x=\frac{2}{\pi}$ using a graphing calculator, (see below) x$\approx$0,69, x$\approx$2.451
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